Since $a + b + c = 1$, we have $a + 1 = 2a + b + c$ and $1 - a = b + c$. Hence we have the equivalent inequality
$$(2a + b + c)\sqrt{2a(b + c)} + (2b + c + a)\sqrt{2b(c + a)} + (2c + a + b)\sqrt{2c(a + b)} \ge 8(ab + bc + ca)$$
From the inequality of arithmetic and geometric mean we get: $2a + b + c \ge 2\sqrt{2a(b + c)}$ and therefore
$$(2a + b + c)\sqrt{2a(b + c)} \ge 2(2a(b + c)) \quad (1)$$
Similarly we have
$$(2b + c + a)\sqrt{2b(c + a)} \ge 2(2b(c + a)) \quad (2),$$
$$(2c + a + b)\sqrt{2c(a + b)} \ge 2(2c(a + b)) \quad (3).$$
Summing by parts (1), (2) and (3) we get the wanted inequality. The equality holds if and only if
$$2a = b + c,\ 2b = c + a,\ 2c = a + b \Leftrightarrow a = b = c = \frac{1}{3}.$$